8. (a) The free-body diagram for the person (shown as an L-shaped block) is shown below. The force
that she exerts on the rock slabs is not directly shown (since the diagram should only show forces
exerted on her), but it is related by Newton’s third law) to the normal forces
N1and
N2exerted
horizontally by the slabs onto her shoes and back, respectively. We will show in part (b) that
N1=N2so that we there is no ambiguity in saying that the magnitude of her push is N2.The
total upward force due to (maximum) static friction is
f=
f1+
f2where (using Eq. 6-1) f1=µs1
N1
and f2=µs2
N2. The problem gives the values µs1 =1.2andµs2 =0.8.
?
mg
N2
-
N1
6
f1
6
f2
(b) We apply Newton’s second law to the xand yaxes (with +xrightward and +yupward and there
is no acceleration in either direction).
N1−N2=0
f1+f2−mg =0
The first equation tells us that the normal forces are equal N1=N2=N. Consequently, from
Eq. 6-1
f1=µs1
N
f2=µs2
N
we conclude that
f1=µs1
µs2f2.
Therefore, f1+f2−mg =0leadsto
µs1
µs2
+1
f2=mg
which (with m=49kg)yieldsf2= 192 N. From this we find N=f2/µs2 = 240 N. This is equal
to the magnitude of the push exerted by the rock climber.
(c) From the above calculation, we find f1=µs1
N= 288 N which amounts to a fraction
f1
W=288
(49)(9.8) =0.60
or 60% of her weight.